$A$ liquid is filled in a container and placed in a room at a temperature of $20^{\circ}C$. When the temperature of the liquid is $80^{\circ}C$,it loses heat at a rate of $60 \, cal/sec$. Find the rate of heat loss when the temperature of the liquid is $40^{\circ}C$ in $cal/sec$.

An ordinary body cools from $4\theta$ to $3\theta$ in $t$ minutes. The temperature of the body after the next $t$ minutes is (Assume Newton's law of cooling and room temperature as $\theta$).

The temperature of a body in air falls from $40^{\circ} \text{C}$ to $24^{\circ} \text{C}$ in $4 \text{ minutes}$. The temperature of the air is $16^{\circ} \text{C}$. The temperature of the body in the next $4 \text{ minutes}$ will be $:$

$A$ hot body,obeying Newton's law of cooling,is cooling down from its peak value $80\,^oC$ to an ambient temperature of $30\,^oC$. It takes $5\,minutes$ to cool down from $80\,^oC$ to $40\,^oC$. How many minutes will it take to cool down from $62\,^oC$ to $32\,^oC$? (Given $\ln 2 = 0.693, \ln 5 = 1.609$)

$A$ body takes $4$ minutes to cool from $100^{\circ}C$ to $70^{\circ}C$. To cool from $70^{\circ}C$ to $40^{\circ}C$ it will take ........ $\text{min.}$ (room temperature is $15^{\circ}C$)

The temperature of a body falls from $62^\circ C$ to $50^\circ C$ in $10$ minutes. If the temperature of the surroundings is $26^\circ C$,the temperature in the next $10$ minutes will become ...... $^\circ C$.